“Clarity about the aims and problems of socialism is of greatest
significance in our age of transition.”

–Albert Einstein

You can access all posts on kinematics on this site by clicking on the label ‘kinematics’ below this post. Or, you may try searching for ‘kinematics’ using the search box at the top of this page.

Today I give you three more multiple choice practice questions with solution:

(1) Tom runs from his school to his home with uniform speed v_{1} and returns to his school with uniform speed v_{2}. The average speed of his round trip is

(a) (v_{1} + v_{2})/2

(b) [(v_{1}^{2} + v_{2}^{2})/2]^{1/2}^{}

(c) (v_{1}v_{2})^{1/2}^{}

(d) v_{1}v_{2}/(v_{1 }+ v_{2})

(e)2v_{1}v_{2}/(v_{1 }+ v_{2})

The average speed is the ratio of the total distance traveled to the total time taken.

If the distance between Tom’s school and his home is s, the total distance for the round trip is 2s and the total time for the round trip is (s/v_{1} + s/v_{2}).

(2) The displacement ‘y’ of a particle thrown vertically down is given by the equation, y= 2t + 5t^{2}, where y is in metre and t is in second. The average velocity during the time interval from 2 s to 2.1 s is

(a) 12.5 ms^{–1}

(b) 16 ms^{–1}

(c) 20.5 ms^{–1}

(d) 22.5 ms^{–1}

(e) 32 ms^{–1}

The velocity ‘v’ of the particle at the instant t is given by

v = dy/dt = 2 + 10t

Therefore, velocities at instants 2 s and 2.1 s are respectively 2 + 10×2 = 22 ms^{–1} and 2 + 10×2.1 = 23 ms^{–1}.

The average velocity v_{average} during during the time interval from 2 s to 2.1 s is given by

v_{average} = (22 + 23)/2 = 22.5 ms^{–1}

[The above question can be answered without using calculus (as in the case of some of the AP Physics B aspirants) like this:

The displacement y is in the form for uniformly accelerated motion in one dimension: y = v_{0}t + ½ at^{2} where v_{0} is the initial velocity and ‘a’ is the acceleration. Therefore, v_{0 }= 2 ms^{–1} and a = 10 ms^{–2}.

The velocity ‘v’ at the instant t is given by v = v_{0} + at = 2 + 10t].

The following question is meant specifically for AP Physics C aspirants:

(3) At the instant t,the position ‘x’ of a particle moving along the x-axis is given by x = 12t^{2} – 2t^{3} where x is in metre and t is in second. What will be the position of this particle when it moves with the maximum speed along the positive x direction?

(a) 32 m

(b) 36 m

(c) 40 m

(d) 48 m

(e) 52 m

The speed v of the particle is given by

v = dx/dt = 24t – 6t^{2}.

When the speed is maximum we have dv/dt = 0

Therefore, 24– 12t = 0 from which t = 2 s.

The position of the particle at 2 seconds will be 12×(2)^{2} – 2×(2)^{3} = 48 – 16 = 32 m (as obtained from x = 12t^{2} – 2t^{3}).

Useful posts in this section can be seen at physicsplus.

The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:

(1) The velocity of escape from the earth’s surface is √(2GM/R) and thework done by earth’s gravitational force to bring unit mass from infinity to a height h from the earth’s surface is GM/(R+h) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass m is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.

(a) R/2

(b) R

(c) 2R

(d) 8R

(e) ∞

The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].

(2) If the body in the above problem is projected up with 20 %of the escape velocity, how high will it rise?

(a) R/24

(b) R/20

(c) R/16

(d) R/8

(e) R/4

The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]^{2} = 0.04 GMm/R.

The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)

[Note that the gravitational potential energy is negative]

Therefore, we have

0.04 GMm/R = GMm/R – GMm/(R+h)

Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]

This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.

(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6R where R is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5R above the surface of the earth will be approximately

(a) 16 hour

(b) 12√2 hour

(c) 8√2 hour

(d) 6√2 hour

(e) 6 hour

According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have

24^{2} α (7R)^{3}since the orbital radius is 6R+R = 7R and the orbital period is 24 hour.

In the case of the research satellite the period T is related to its orbital radius by

T^{2} α (3.5R)^{3} since its orbital radius is 2.5R+R = 3.5R^{}

The following questions are meant solely for AP Physics C aspirants:

(4) Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by

(a) 1%

(b) 1.5%

(c) 2%

(d) 2.5%

(e) 3%

We have from Kepler’s third law, T^{2} α R^{3} where T and R are the orbital period and orbital radius respectively.

Therefore, T = kR^{3/2} where k is the constant of proportionality.

Taking logarithms, ln T = ln k + (3/2) ln R

Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.

In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].

(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will

(a) remain unchanged

(b) decrease by 2%

(c) increase by 2%

(d) increase by 1%

(e) decrease by 1%

The orbital speed v of a satellite moving around the earth in an orbit of radius r is given by

v = √(GM/r) where G is the gravitational constant and M is the mass of the earth.

[You can easily obtain this by equating the centripetal force to the gravitational pull: mv^{2}/r = GMm/r^{2} where m is the mass of the satellite].

This equation shows that the orbital speed will decrease when the orbital radius is increased.

Since G and M are constants, we have

dv/v = – ½ dr/r

[You will get this by taking logarithms and by differentiating, as we did in question No.4].

This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the decrease in the orbital speed because of the increase in orbital radius.

Since the orbital radius of the satellite is increased by 2%, the orbital speed is decreased by 1% [Option (e)].

(6) A communication satellite of mass m is placed initially in a temporary orbit of radius r_{1} around the earth_{.} How much work is to be done to shift it from this orbit to a permanent orbit of greater radius r_{2}?_{ }(Acceleration due to gravity at the sea level = g. Radius of the earth = R)

(a) mgR^{2}(r_{2}–r_{1})/ r_{2}r_{1}

(b) mgR (r_{2}–r_{1})/ 2r_{2}

(c) mgR^{2}(r_{2}–r_{1})/ 2r_{2}r_{1}

(d) mgR^{2}(r_{2}–r_{1})/ 2r_{2}r_{1}

(e) 2mgR^{2}(r_{2}–r_{1})/ r_{2}r_{1}

Total energy (K.E. + P.E.) of a satellite of mass m in an orbit of radius r is – GMm/2r where M is the mass of the earth.

Work (W) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.

Therefore, W = – GMm/2r_{2} – (– GMm/2r_{1})

[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].

Thus W = (GMm/2)(1/r_{1 }– 1/r_{2}) = (GMm/2)[(r_{2}–r_{1})/ r_{2}r_{1}]

Since g = GM/R^{2} (as you can obtain from mg = GMm/R^{2}), we have

The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:

(1) The velocity of escape from the earth’s surface is √(2GM/R) and thework done by earth’s gravitational force to bring unit mass from infinity to a height h from the earth’s surface is GM/(R+h) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass m is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.

(a) R/2

(b) R

(c) 2R

(d) 8R

(e) ∞

The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].

(2) If the body in the above problem is projected up with 20 %of the escape velocity, how high will it rise?

(a) R/24

(b) R/20

(c) R/16

(d) R/8

(e) R/4

The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]^{2} = 0.04 GMm/R.

The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)

[Note that the gravitational potential energy is negative]

Therefore, we have

0.04 GMm/R = GMm/R – GMm/(R+h)

Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]

This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.

(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6R where R is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5R above the surface of the earth will be approximately

(a) 16 hour

(b) 12√2 hour

(c) 8√2 hour

(d) 6√2 hour

(e) 6 hour

According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have

24^{2} α (7R)^{3}since the orbital radius is 6R+R = 7R and the orbital period is 24 hour.

In the case of the research satellite the period T is related to its orbital radius by

T^{2} α (3.5R)^{3} since its orbital radius is 2.5R+R = 3.5R^{}

The following questions are meant solely for AP Physics C aspirants:

(4) Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by

(a) 1%

(b) 1.5%

(c) 2%

(d) 2.5%

(e) 3%

We have from Kepler’s third law, T^{2} α R^{3} where T and R are the orbital period and orbital radius respectively.

Therefore, T = kR^{3/2} where k is the constant of proportionality.

Taking logarithms, ln T = ln k + (3/2) ln R

Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.

In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].

(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will

(a) remain unchanged

(b) decrease by 2%

(c) increase by 2%

(d) increase by 1%

(e) decrease by 1%

The orbital speed v of a satellite moving around the earth in an orbit of radius r is given by

v = √(GM/r) where G is the gravitational constant and M is the mass of the earth.

[You can easily obtain by equating the centripetal force to the gravitational pull: mv^{2}/r = GMm/r^{2} where m is the mass of the satellite].

This equation shows that the orbital speed will decrease when the orbital radius is increased.

Since G and M are constants, we have

dv/v = – ½ dr/r

[You will get this by taking logarithms and by differentiating, as we did in question No.4].

This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the decrease in the orbital speed because of the increase in orbital radius.

Since the orbital radius of a the satellite is increased by 2%, the orbital speed is decreased by 1% [Option (e)].

(6) A communication satellite of mass m is placed initially in a temporary orbit of radius r_{1} around the earth_{.} How much work is to be done to shift it from this orbit to a permanent orbit of greater radius r_{2}?_{ }(Acceleration due to gravity at the sea level = g. Radius of the earth = R)

(a) mgR^{2}(r_{2}–r_{1})/ r_{2}r_{1}

(b) mgR (r_{2}–r_{1})/ 2r_{2}

(c) mgR^{2}(r_{2}–r_{1})/ 2r_{2}r_{1}

(d) mgR^{2}(r_{2}–r_{1})/ 2r_{2}r_{1}

(e) 2mgR^{2}(r_{2}–r_{1})/ r_{2}r_{1}

Total energy (K.E. + P.E.) of a satellite of mass m in an orbit of radius r is – GMm/2r where M is the mass of the earth.

Work (W) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.

Therefore, W = – GMm/2r_{2}– (– GMm/2r_{1})

[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].

Thus W = (GMm/2)(1/r_{1 }– 1/r_{2}) = (GMm/2)[(r_{2}–r_{1})/ r_{2}r_{1}]

Since g = GM/R^{2} (as you can obtain from mg = GMm/R^{2}), we have

A free response practice question involving waves and electromagnetism was given to you in the post dated 30^{th} October 2009. As promised, I give below a model answer along with the question:

Your teacher wants you to measure the frequency of a sinusoidal alternating current, making use of the resonance condition in a vibrating wire segment. She gives you a sonometer (fig.) with a steel wire of linear density m, standard weights, a small electromagnet for vibrating the wire, an AC source of variable output voltage but of fixed frequency (which is to be measured), connecting wires and metre scale. Now answer the following questions:

(a) Show in a diagram how you will arrange the given items for the measurement of the frequency of the given AC source.

(b) You are given that the frequency of vibration of the sonometer wire segment of length ℓ in the s^{th}mode is (s/2ℓ)√(T/m) where T is the tension.

(i) Now explain briefly how you will proceed to measure the frequency of the alternating current making use of the vibration of the sonometer wire segment in the fundamental mode.

(ii) Write down the expression for the frequency of the wire.

(ii) How is the frequency of AC related to the frequency of vibration of the sonometer wire? Justify your answer.

(c) Your teacher now wants you to use a brass wire of linear density μ instead of the steel wire in the sonometer. Instead of the electromagnet she gives you a sufficiently strongpermanent bar magnet.

Show in a diagram how you will arrange the given items for the measurement of the frequency of the given AC source.

(d) (i) Explain briefly how you will proceed to measure the frequency of the alternating current in this case.

(ii) How is the frequency of AC related to the frequency of vibration of the sonometer wire in this case? Justify your answer.

(a) The arrangement for measuring the frequency is shown in the following figure in which the electromagnet is shown connected to the AC source:

(b)(i) By suspending a suitable mass M, the steel wire in the sonometer is kept under tension T (= Mg). The electromagnet energized by the AC source is held above the steel wire segment AB in the sonometer. The wire oscillates because of the periodic attraction by the electromagnet. The length of the wire segment AB is adjusted by moving the knife edge A or B of the sonometer so that the wire segment is at resonance. This condition is indicated by the maximum amplitude of vibration of the wire segment, which can be conveniently judged by the maximum flutter of a small paper rider placed at the middle of the wire. The minimum length ℓ of the wire segment for attaining the resonance condition is measured using a metre scale.

(ii) The frequency of vibration of the sonometer wire segment of length ℓ in the s^{th}mode is (s/2ℓ)√(T/m) where T is the tension and m is the linear density of the wire. In the fundamental mode, s = 1. Therefore, the frequency of vibration of the sonometer wire segment = (1/2ℓ)√(T/m).

(iii) Since the steel wire is attracted towards the electromagnet during the negative as well as positive peak of the alternating current flowing through the electromagnet, the frequency of vibration of the wire is twice the frequency of the AC.

Therefore, frequency of the AC = (1/4ℓ)√(T/m).

(c) For the measurement of the frequency of AC using the vibration of the brass wire, the alternating current is passed through the brass wire (in the sonometer) by connecting it in series with the AC source. The bar magnet is arranged near the wire so that the axis of the magnet is horizontal and at right angles to the wire. The arrangement is shown in the following figure

(d)

(i) The current carrying wire of the sonometer is located in the magnetic field produced by the bar magnet and hence it will experience magnetic forces. Since the current is alternating, the force is periodic and the wire will vibrate. Keeping the wire under a suitable tension T_{1 }(= M_{1}g), the length of the wire segment AB is adjusted by moving the knife edge A or B of the sonometer so that resonance is attained. The minimum length of the wire segment ℓ_{1} (corresponding to the fundamental mode of vibration) is measured. The frequency of vibration of the wire is (1/2ℓ_{1} )√(T_{1}/μ).

(ii) The current carrying brass wire is located in the magnetic field of the bar magnet and hence feels a magnetic force. The wire is pulled up during one half cycle of AC and pushed down during the next half cycle since the direction of the current in the wire changes. This means that the frequency of vibration of the wire is the same as that of the AC passing through the wire. Therefore, frequency of AC = (1/2ℓ_{1} )√(T_{1}/μ).